Question: $f(x, y) = xy - e^{xy}$ What is the Laplacian of $f(x, y)$ ? $\Delta f = $
Solution: The Laplacian of a scalar field $f$ is the sum of each of its second partial derivatives. $\Delta f = \dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2}$ [What does that triangle mean?] Let's find the second partial derivatives of $f$ ! $\begin{aligned} f_{xx} &= \dfrac{\partial}{\partial x} \left[ \dfrac{\partial f}{\partial x} \right] = \dfrac{\partial}{\partial x} \left[ y - ye^{xy} \right] = -y^2e^{xy} \\ \\ f_{yy} &= \dfrac{\partial}{\partial y} \left[ \dfrac{\partial f}{\partial y} \right] = \dfrac{\partial}{\partial y} \left[ x - xe^{xy} \right] = -x^2e^{xy} \end{aligned}$ The Laplacian is $\Delta f = f_{xx} + f_{yy}$. Therefore: $\Delta f(x, y) = -e^{xy} \left( x^2 + y^2 \right)$